{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

topics_11_3_11_4_11_5 - Topics 11.3 11.4 11.5 11.3 Polar...

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Topics 11.3, 11.4, 11.5 11.3 Polar Coordinates Polar coordinate system introduced by Newton : It has a pole that is the origin O of a Cartesian coordinate system . It has a polar axis that coincides with the right-half of the x -axis in the Cartesian coordinate system, starting from the pole O. If P 6 = O is a point on the plane and r is the distance from the pole O to the point P and θ is the angle between the polar axis and the line OP assuming that the counter-clockwise direction is a positive direction, then ( r , θ ) are the polar coordinates of the point P . We agree the pole O to be represented by (0 , θ ) for any value of θ ( r = 0 at the pole O ). Now suppose that the point P is tracing a curve C . Then r = f ( θ ) , α θ β is a representation of the curve C called polar representation. In fact, it is equivalent to the following parametric equations of the curve C : x = f ( θ ) cos( θ ) , y = f ( θ ) sin( θ ) , α θ β. The equation r = f ( θ ) , α θ β is called a polar equation of a curve. It is tracing a curve. The angle θ is a parameter in the polar representation of a curve. 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Important remarks for plots! (1) We extend the meaning of polar coordinates ( r, θ ) to r negative, agreeing that the points ( - r , θ ) and ( r , θ ) are on the same line, equidistant to O , but opposite with respect to O . In other words, ( - r , θ ) represents a point with polar coordi- nates ( r , θ + π ) . (2) A polar curve (a curve given by a polar equation) is symmetric about the x - axis if r ( - θ ) = r ( θ ). (3) A polar curve is symmetric about the y - axis if r ( π - θ ) = r ( θ ). (4) A polar curve is symmetric with respect to the origin if r ( π + θ ) = r ( θ ), i.e., if replacing r by - r we obtain an equivalent polar equation. Problem 1. Sketch the region in the plane consisting of points whose polar coordinates satisfy 1 r < 3 , - π/ 3 < θ 2 π/ 3 . Problems 31, 35, 36, 37, 41, 42/ page 684. Sketch the curve with the given polar equation. 31 . r = sin( θ ); 35 . r = θ, θ 0; 36 . r = ln( θ ) , θ 1 37 . r = 4 sin(3 θ ); 41 . r = 1 - 2 sin( θ ); 42 . r = 2 + sin( θ ) 2
Image of page 2
By using CAS MAPLE code plot ([[ 1 * cos ( t ) , 1 * sin ( t ) , t = - Pi / 3 .. 2 * Pi / 3 ] , [ 3 * cos ( t ) , 3 * sin ( t ) , t = - Pi / 3 .. 2 * Pi / 3 ] , [ r * cos ( - Pi / 3 ) , r * sin ( - Pi / 3 ) , r = 1 .. 3 ] , [ r * cos ( 2 * Pi / 3 ) , r * sin ( 2 * Pi / 3 ) , r = 1 .. 3 ]] , color = [ black , red , red , black ]); we plot the region to verify the result of Problem 1. 3
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Plots of Problems 31, 35, 36 by using CAS MAPLE polar- plot. We give an explicit form of the CAS MAPLE polar-plot for Problem 31: r = sin( θ ). For convenience we use the notation t instead of θ : polarplot ( sin ( t ) , t = 0 .. 2 * Pi ); Plots of Problems 37, 41, 42 by using CAS MAPLE polar-plot. 4
Image of page 4
Problems 15-20, page 684. Identify the curve by finding a Cartesian equation for the curve: 15 . r = 2; 16 . r cos( θ ) = 1; 17 . r = 3 sin( θ ); 18 . r = 2 sin( θ ) + 2 cos( θ ); 19 . r = csc( θ ); 20 . r = tan( θ ) sec( θ ) . Hint. Use x = r cos( θ ) , y = r sin( θ ). Solution of Problem 18. Multiplying both sides by r we get r 2 = 2 r sin( θ )+2 r cos( θ ) x 2 + y 2 = 2 y +2 x ( x - 1) 2 +( y - 1) 2 = 2 and this is a circle of radius 2, centered at (1 , 1).
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern